Question

Sixteen players $S_1,S_2,...,S_{16}$ play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pair. Assuming that all the players are to equal strength. The probability that the players $S_1$ is among the eight winners is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2}$

Since all players are of equal strength, with whom $S_1$ is paired, will have no bearing on the probability of his winning.
The number of ways of choosing $8$ winners out of $16$ is ${}^{16}{C}_8$
The number of ways of choosing $S_1$ and $7$ other winners out of $15$ is ${}^{15}{C}_7$
$\therefore$ Probability that $S_1$ will win $=\frac{{}^{15}{C}_7}{{}^{16}{C}_8}=\frac{15!}{7!8!}\frac{8!8!}{16!}=\frac{1}{2}$